Explanation Approach: If a system of 3 linear equations in the 3 variable x , y , z is equivalent to the matrix equation D X = B , and if D ≠ 0 , then its solutions are x = | D x | | D | , y = | D y | | D | , z = | D z | | D | Here, D x , D y , a n d D z are the matrices obtained by replacing the 1 st , 2 nd , and 3 rd column of D by the n × 1 matrix B respectively. Calculation: The system of equations { 2 x − z = 14 3 x − y + 5 z = 0 4 x + 2 y + 3 z = − 2 is equivalent to the matrix equation D X = B . Here, D = [ 2 0 − 1 3 − 1 5 4 2 3 ] , X = [ x y z ] , and B = [ 14 0 − 2 ] . | D | = | 2 0 − 1 3 − 1 5 4 2 3 | = 2 ⋅ ( − 13 ) − 0 − 1 ⋅ 10 = − 36 | D x | = | 14 0 − 1 0 − 1 5 − 2 2 3 | = 14 ⋅ ( − 13 ) − 0 ⋅ 10 + ( − 1 ) ⋅ ( − 2 ) = − 180

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ISBN: 8220100548512

Author: Watson

Publisher: CENGAGE L

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Chapter 11.CT, Problem 19CT

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The solution of the system {2x −z=143x− y+5z= 04x+ 2y+3z=−2 using Cramer’s rule.

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Chapter 11.CT, Problem 18CT

Chapter 11.CT, Problem 20CT

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The solution of the system { 2 x − z = 14 3 x − y + 5 z = 0 4 x + 2 y + 3 z = − 2 using Cramer’s rule.

Solution Summary: The author explains how the solution of the system c2x, , and is equivalent to the matrix equation DX=B.

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To find:

The solution of the system {2x −z=143x− y+5z= 04x+ 2y+3z=−2 using Cramer’s rule.

Want to see the full answer?

Chapter 11 Solutions

The solution of the system { 2 x − z = 14 3 x − y + 5 z = 0 4 x + 2 y + 3 z = − 2 using Cramer’s rule.

Solution Summary: The author explains how the solution of the system c2x, , and is equivalent to the matrix equation DX=B.

Videos

To find: The solution of the system {2x −z=143x− y+5z= 04x+ 2y+3z=−2 using Cramer’s rule.

Want to see the full answer?

Chapter 11 Solutions

To find:

The solution of the system {2x −z=143x− y+5z= 04x+ 2y+3z=−2 using Cramer’s rule.